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This question already has an answer here:

Consider the polynomial
$$p(x) = a_0 + a_1x + a_2x^2 + · · · + a_nx^n$$ where $a_0, a_1, . . . , a_n ∈ \Bbb Z$. Show that if $p(x_i) = 7$ for 4 distinct integers $x_0, x_1, x_2, x_3$, then $p(z) \neq 14$ for any $z ∈ \Bbb Z$.

How do I start this question?

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marked as duplicate by Bill Dubuque polynomials Dec 14 '16 at 18:49

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Let $q(x)=p(x)-7$. Now the problem becomes:

Show that if $q(x_i)=0$ for $4$ distinct integers $x_0,x_1,x_2,x_3,$ then $q(z) \ne 7$ for any $z \in \mathbb Z$

But if $q(x_i) = 0$ for $i=0,\ldots,3,$ then

$$q(x)=(x-x_0)(x-x_1)(x-x_2)(x-x_3)r(x)$$

for some integer polynomial $r(x)$. Suppose $q(z) = 7$ for some $z \in \mathbb Z$. Can you see the contradiction?

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  • $\begingroup$ Sorry, I don't get it $\endgroup$ – user286826 Jan 19 '16 at 1:01
  • $\begingroup$ If $q(z)=7$, then what possible values can $z-x_0,\ldots,z-x_3$ have? Remember, they are all different. $\endgroup$ – TonyK Jan 19 '16 at 1:03
  • $\begingroup$ I get it. 7 is prime so $z−x_0,…,z−x_3$ don't exist. $\endgroup$ – user286826 Jan 19 '16 at 1:09
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    $\begingroup$ Right! You can have at most three distinct integers whose product divides $7$. For instance, $1, 7, -1$. $\endgroup$ – TonyK Jan 19 '16 at 1:19

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